For 16, Q and T are in cages. Based on the rules of the game and the odd number of Parakeets, you know that a pair of Parakeets MUST be exhibited (odd number of parakeets = impossible to house all males in either cage without braking rules). Now that you have established that a Male/Female Parakeet must always be exhibited, since you don't have T available as a pair for a female parakeet you must have W as the female Parakeet. Based on the last rule you deduce that the pair is RW, since if you have W you cannot have S (the only other available option to compliment the pair).

For 18, you know S is exhibited. Based on the last rule again, since you have S you cant have W as a pair (pair would be ST). Since W is the only other available pair for male Parakeets you know that Q and R must reside in cages because they dont have an available pair to be grouped with.

It is much easier explaining these in person but I hope this clears a few things up.

Just woke up this morning and looked at the question again. The first line of the game says "The breeder exhibits pairs of birds consisting of one male and one female OF THE SAME KIND." I have been matching male/female birds on exhibition regardless of their kind.